Question

The circumference of a sphere was measured to be 90 cm with a possible error of 0.5 cm.

Answer 1

Answer:

Error in the sphere's surface: 29 $cm^2$  and relative error in surface measure: 0.011

Error in the sphere's volume: 205 $cm^3$ and relative error in the volume measure: 0.017

Step-by-step explanation:

(a)

The measured length (l) of the circumference is 90 cm with an error of 0.5 cm, that is:

$l=2\,\pi\,R=90\,cm\\R=\frac{90}{2\,\pi} \,cm=\frac{45}{\pi} \,cm=14.3239\,\,cm$

and with regards to the error:

$dl=0.5 \, cm\\dl=2\,\pi\,dR\\dR=\frac{dl}{2\,\pi} =\frac{1}{4\,\pi} cm = 0.0796\,cm$

then when we use the formula for the sphere's surface, we get:

$S=4\,\pi\,R^2\\dS=4\,\pi\,2\,R\,(dR)\\dS=8\,\,\pi\.(\frac{45}{\pi} \,\,cm)\,(\frac{1}{4\pi}\,cm) =\frac{90}{\pi} \,\,cm^2\approx \,29\,cm^2$

Then the relative error in the surface is:

$\frac{dS}{S} =\frac{90/\pi}{4\,\pi\,R^2} =\frac{1}{90} =0.011$

(b)

Use the formula for the volume of the sphere:

$V=\frac{4\,\pi}{3} R^3\\dV=\frac{4\,\pi}{3}\,3\,R^2\,(dR)=4\,\pi\,R^2\,(\frac{1}{4\pi}) \,cm=(\frac{45}{\pi})^2 \,\,cm^3\approx 205\,\,cm^3$

Then the relative error in the volume is:

$\frac{dV}{V} =\frac{205}{12310.5} \approx 0.017$